IFPOS

AcronymDefinition
IFPOSInternational Federation of Pediatric Orthopaedic Societies
References in periodicals archive ?
B is an IFOS in (Y, [sigma]), and B is an IFPOS in Y, since B [subset or equal to] int(cl(B)) = B.
Now [([P.sub.i] * f).sup.-1] ([B.sub.i]) = [f.sup.-1]([P.sup.-1.sub.i]([B.sub.i])), since f is IF pre-[beta]-irresolute and [P.sup.-1.sub.i] ([B.sub.i]) is IF[beta]OS, [f.sup.-1]([P.sup.-1.sub.i] ([B.sub.i])) is IFPOS in X.
Since [f.sub.1] and [f.sub.2] are IF pre-[beta]-irresolute, [([f.sub.1] x [f.sub.2]).sup.-1] (C) is an IFPOS in [X.sub.1] x [X.sub.2] and hence [f.sub.1] x [f.sub.2] is IF pre-[beta]-irresolute function.
A mapping f : X [right arrow] Y from an IFTS X into an IFTS Y is IF pre-[beta]-irresolute if and only if for each IFP[p.sub.([alpha], [beta])] in X and IF[beta]OSB in Y such that f([p.sub.([alpha], [beta])]) [member of] B, there exists an IFPOS A in X such that [p.sub.([alpha], [beta])] [member of] A and f(A) [subset or equal to] B.
Then A is an IFPOS in X which containing IFP[p.sub.([alpha], [beta])] and f(A) = f([f.sup.-1](B)) [subset or equal to] B.
According to assumption there exists IFPOS A in X such that [p.sub.([alpha], [beta])] [member of] A and f(A) [subset or equal to] B.
Since f is IF pre-[beta]-irresolute, [f.sup.-1]([g.sup.-1](B)) is IFPOS in X.
Since f is IF pre-[beta]-irresolute, [f.sup.-1]([g.sup.-1](B)) is IFPOS in X which implies g * f is IF pre continuous.